Question

Решите уравнения:
x^2+3x=25+3x
(x-3)(x+3)-4=6
(x-13)^2=3

Answer 1

1)

$x {}^{2} + 3x = 25 + 3x$

Равные члены взаимно уничтожаются

$x {}^{2} = 25 \\ x = ±5 \\ x1 = - 5⠀⠀⠀x2 = 5$

2)

$(x - 3)(x + 3) - 4 = 6 \\ x {}^{2} - 9 - 4 = 6 \\ x {}^{2} - 13 = 6 \\ x {}^{2} = 6 + 13 \\ x {}^{2} = 19 \\ x = ± \sqrt{19} \\ x1 = - \sqrt{19} ⠀⠀⠀x2 = \sqrt{19}$

3)

$(x - 13) {}^{2} = 3 \\ {x}^{2} - 26x + 166 = 0 \\D= ( - 26) {}^{2} - 4 \times 1 \times 166 = 12 \\ x1 = \frac{ - ( - 26) + \sqrt{12} }{2 \times 1} = \frac{26 + 2 \sqrt{3} }{2} = \frac{2(13 + \sqrt{3)} }{2} = 13 + \sqrt{3} \\ x2 = \frac{ - ( - 26) - \sqrt{12} }{2 \times 1} = \frac{26 - 2 \sqrt{3} }{2} = \frac{2(13 - \sqrt{3)} }{2} = 13 - \sqrt{3} \\ Ответ:x1 = 13 + \sqrt{3} ⠀⠀⠀x2 = 13 - \sqrt{3}$